Hazardous materials are grouped into classes identifying their similarities in composition and structure.
<h3>Why hazardous materials are grouped into classes?</h3>
The hazardous materials are grouped into classes in order to tell us about the severity of hazard and it is done on the basis of similarity in composition.
So we can conclude that hazardous materials are grouped into classes identifying their similarities in composition and structure.
Learn more about hazardous here: brainly.com/question/7310653
Answer:
The molar solubility of lead bromide at 298K is 0.010 mol/L.
Explanation:
In order to solve this problem, we need to use the Nernst Equaiton:
![E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}](https://tex.z-dn.net/?f=E%20%3D%20E%5E%7Bo%7D%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D)
E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.
At equilibrium, E = 0, therefore:
![E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\](https://tex.z-dn.net/?f=E%5E%7Bo%7D%20%20%3D%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%20%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D%20%5C%5C%5C%5Clog%20%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D%20%3D%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%20%5C%5C%5C%5Clog%5Bred%5D%20%3D%20%20log%5Box%5D%20-%20%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%5C%5C%5C%5C%5Bred%5D%20%3D%2010%5E%7B%20log%5Box%5D%20-%20%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%7D%20%5C%5C%5C%5C%5Bred%5D%20%3D%2010%5E%7B%20log0.733%20-%20%20%5Cfrac%7B2x5.45x10%5E%7B-2%7D%20%20%7D%7B0.0591%7D%7D%5C%5C%5C%5C)
[red] = 0.010 M
The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.
Nitrogen could form 3 covalent bonds if each of its unpaired electrons participates in one bond.
Nitrogen atom forms 3 bonds based on octet rule, because it has 5 valence electrons. That means it needs 3 bonds or three more electrons.
<h3>Further Explanation;</h3><h3>Chemical bond </h3>
- A bond is a type of force that is formed between atoms of different through the sharing or transfer of electrons.
<h3>Octet rule</h3>
- According to the octet rule for an atom to be stable it must have maximum number of electrons in its outermost energy level. Therefore an atom with four electrons requires four more electrons to attain stability.
<h3>Types of chemical bonds.</h3><h3>Covalent bond </h3>
- This is a type of bond that is formed between non-metal atoms. It is formed as a result of sharing electrons between non-metal atoms involved.
- When atoms involved contribute equal number of electrons to the bond formation, the type of bond is known as covalent bond
- A covalent bond may be a dative covalent bond, when the shared electrons come from one atom.
<h3>Ionic bond </h3>
- This is a type of bond that occurs between metal ions and non-metal ions. Ionic bond occurs as a result of transfer of electrons from one metal atom to another non-metal atom.
- After the transfer of electrons, metal atom loses electron to form a cation while the non-metal atom gains electrons to form an anion.
<h3>
Other types of chemical bonds include;</h3>
- Hydrogen bonds
- Metallic bonds
- Dipole-dipole interactions, etc.
Keywords: Chemical bond, covalent bond, atom
<h3>Learn more about:</h3>
Level: High school
Subject: Chemistry
Topic: structure and bonding
Sub-topic: Covalent bond
Answer:
The percentage of N in the compound is 0.5088
Explanation:
Mass of compound = 8.75 mg = 8.75×1000 = 8750 g
Mass of N2 = number of moles of N2 × MW of N2 = 1.59 × 28 = 44.52 g
% of N in the compound = (mass of N2/mass of compound) × 100 = (44.52/8750) × 100 = 5.088×10^-3 × 100 = 0.5088
Explanation:
Flourine has atomic number of 9 and hence 9 electrons in its neutral state. The full electronic configuration is given as;
1s2 2s2 2p5
Carbon has atomic number of 6 and hence 6 electrons in it's neutral state. The noble gas notation as the following format;
[closest noble gas before the element] remaining electrons
The nearest noble gas to carbon is Helium, the noble gas notation is given as;
[He] 2s4
